jmc

We can see that $$\begin{array}{rl}{1}_2+10_2+100_2+\cdots +100000000_2& =111111111_2\\ & =1000000000_2-1\\ & =2^9-1.\end{array}$$We can factor $2^9-1=8^3-1$ as a difference of cubes to make our task easier: $$8^3-1=(8-1)(8^2+8+1)=7\cdot 73.$$Since $\boxed{73}$ is prime, it is the largest prime divisor of the sum.